Let $S,T,L$ be the intersections of $AD,BC$ and $AC,BD$ and $PM,QN$ respectively. Using desargues theorem for triangles $PMD$ and $QNC$ we should prove $S,T,L$ are collinear. Which is obvious using Pascal theorem for $APMBQN$.So we are done.$\square$

Just apply Pascal's theorem for $QPANMB$

Another way to solve the problem is by simple angle chasing prove that quadrilaterals $CDMN$ and $CDPQ$ are cyclic. Let $\omega_1$ be the circumcircle of $MNPQ$; let $\omega_2$ be the circumcircle of $CDMN$; let $\omega_3$ be the circumcircle of $CDPQ$. Then $CD$ is the radical axis of $\omega_2$ and $\omega_3$; $PQ$ is the radical axis of $\omega_3$ and $\omega_1$; $MN$ is the radical axis of $\omega_1$ and $\omega_2$. So $MN$, $CD$, $PQ$ are concurrent on the radical center